The cone pictured above has a radius of 1.5cm and a slant height of 3.5cm. Letting $\alpha=\arcsin\left(\frac{3}{7}\right)$ and $h=3.5\cos\alpha$, I have parametrized the cone in two ways:
$r(t,u)=\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix} \left(\frac{3}{2h}\right)t\cos u \\ \left(\frac{3}{2h}\right)t\sin u\\t \end{bmatrix},\ t\in[0,h],\ u\in[0,2\pi]$
$r(\rho,\varphi,\theta)=\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix} \rho\cos\theta\sin\varphi \\ \rho\sin\theta\sin\varphi \\ \rho\cos\varphi \end{bmatrix},\ \rho\in[0,h\sec\varphi],\ \varphi\in[0,\alpha],\ \theta\in[0,2\pi]$
With $r(t,u)$, and noting that $r_t=\frac{\partial}{\partial t}r(t,u)$, I can find the surface area, $SA$, of the cone by
$SA=\displaystyle\int_0^{2\pi}\int_0^h ||r_t\times r_u||\ dt\ du$.
With $r(\rho,\varphi,\theta)$, I can find the volume, $V$, of the cone by taking the vectors $r_{\rho}$, $r_{\varphi}$, and $r_{\theta}$, putting them together to form a matrix $J_r$ (the Jacobian of $r(\rho,\varphi,\theta)$) and letting $G=(J_r)^TJ_r$, I take the $\det(G)$ and my volume is given by
$V=\displaystyle\int_0^{2\pi}\int_0^{\alpha}\int_0^{h\sec\varphi}\sqrt{\det(G)}\ d\rho\ d\varphi\ d\theta$.
What I hope someone can inform me on, however, is how to find the volume of the cone using $r(t,u)$ rather than $r(\rho,\varphi,\theta)$. I've scowered the internet on how to find the volume of a parametric surface, but, to no avail. I would greatly appreciate any help!
Best Answer
In the first case, you are parametrizing the lateral surface of the cone and in the second, you are defining the interior points too. In general, it is unnecessary to parametrize the surface to find the volume bound. But if you were given a parametrized surface with appropriate bounds, you could of course use it to find the volume, just like you would do if for example, cone was given as $z \tan \alpha = \sqrt{x^2+y^2}, 0 \leq z \leq h$. So in this case,
$r(t,u)=\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix} \left(\frac{3t}{2\sqrt{10}}\right)\cos u \\ \left(\frac{3t}{2\sqrt{10}}\right)\sin u\\t \end{bmatrix},\ t\in[0,h],\ u\in[0,2\pi]$
(please note there is a mistake in your parametrization)
$ \displaystyle V = \int_0^{2\pi} \int_0^h \int_0^{3t / (2 \sqrt{10})} r ~ dr ~ dt ~ du$
See the variable $r$ and its bounds for every $t$ and $u$, which is radius of the cross section of the cone at height $t$.